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Let $x=(x_1,\ldots,x_n)\in\mathbb{R}^n$. Can one find a polynomial $p$ (of arbitrary degree) in the coordinates of $x$ such that $p(x)\geq 0$ if and only if $x$ is an element of the positive orthant $\{x:x_1\geq 0,\ldots,x_n\geq0\}$?

If so, what is the polynomial of minimal degree? If not, can one find a "best" approximation to such a function with polynomials?

A simple remark is that since $x\mapsto p(x)$ is continuous, we know that it vanishes on the coordinate axes and must be of the form $p(x)=x_1\cdots x_n q(x)$.

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As stated, the answer is trivially no, since as you observe, the condition forces $p$ to also vanish on the coordinate hyperplanes, which are not contained in the positive orthant. If you modify the question to say that $p(x)>0$ iff $x$ is in the open positive orthant, the answer is still no (for $n\geq 2$). As you observe, we can write $p(x)=x_1\dots x_nq(x)$ for some $q$. But then $q$ must change sign when you change between any two adjacent orthants that are not the positive orthant (since $x_1\dots x_n$ changes sign but $p$ does not), so $q$ is also divisible by $x_1\dots x_n$. We can thus write $p(x)=x_1^2\dots x_n^2r(x)$ where $r$ is also positive only in the positive orthant, and so by induction on degree no such $p$ can exist.

I'm not sure what kind of "approximation" you're looking for, but here's a fairly strong sense in which no approximation can exist with nice uniformity properties as $x$ gets large. Let $p$ be any nonconstant polynomial in $n$ variables and consider the single variable polynomials $q_x(t)=p(tx)$. For almost every $x$ (i.e., for $x$ not lying in a proper subvariety), $q_x$ will have the same degree as $p$. In particular, if $p$ has even degree then for almost every $x$, $p(tx)$ will have the same sign as $t\to\infty$ and $t\to-\infty$, and if $p$ has odd degree $p(tx)$ will almost always have opposite sign as $t\to \infty$ and $t\to -\infty$. On the other hand, a function that is positive in the positive orthant and negative in all the other orthants does not behave like either an even or an odd degree polynomial in this respect. In particular, any nonconstant polynomial "approximation" must be large and have the wrong sign on (asympotically) at least $1/2^n$ of the volume of large balls around the origin.

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  • $\begingroup$ Thanks! I was really surprised by the last part about no good polynomial approximations. $\endgroup$
    – fact
    Commented Dec 2, 2014 at 20:06
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No. Let $x_1$ divide $p(x)$ with multiplicity $k$. Since $p$ switches sign when crossing from the $+++\cdots+$ quadrant to the $-+++\cdots+$ quadrant, $k$ is odd. But, since $p$ doesn't switch sign when crossing from $+-++\cdots+$ to $--++\cdots+$, we see that $k$ is also even. Contradiction.

I feel like this question has appeared here before.

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  • $\begingroup$ It does sound familiar, but all I could find was this question on math.se: math.stackexchange.com/questions/966633/… $\endgroup$ Commented Dec 2, 2014 at 17:47
  • $\begingroup$ Thanks for the elegant answer. I agree, it seems such a natural question that it must have been asked before but I could only find things about the "Positivstellensatz" which seems to not directly apply to my question (but then again, I have 0 knowledge of algebraic geometry). $\endgroup$
    – fact
    Commented Dec 2, 2014 at 20:10
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The correct approach is that of hyperbolic polynomials.

If $p=:p_1$ is homogeneous, hyperbolic in a direction $\bf e$, then we define the successive derivatives $p_2={\bf e}\cdot\nabla p$, then $p_{j+1}={\bf e}\cdot\nabla p_j$. The forward cone $\Gamma_p$ is defined as the connected component of $\bf e$ in the complement of $\{p(x)=0\}$. Then a theorem of Garding says that $\Gamma_p$ is the set defined by the equalities $$p_1(x)\ge0,\ldots,p_d(x)\ge0.$$

In your case, the polynomial is $p(x)=x_1\cdots x_n$, with $d=n$. And $\bf e$ is any vector taken in the positive orthant. If you choose ${\bf e}=(1,\ldots,1\}$, you obtain the characterization $$(x\ge0)\Longleftrightarrow(\sigma_1(x)\ge0,\,\ldots,\sigma_n(x)\ge0).$$

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